Solve for $x$ : $ 3|x - 1| - 4 = -3|x - 1| + 7 $
Answer: Add $ {3|x - 1|} $ to both sides: $ \begin{eqnarray} 3|x - 1| - 4 &=& -3|x - 1| + 7 \\ \\ { + 3|x - 1|} && { + 3|x - 1|} \\ \\ 6|x - 1| - 4 &=& 7 \end{eqnarray} $ Add ${4}$ to both sides: $ \begin{eqnarray} 6|x - 1| - 4 &=& 7 \\ \\ { + 4} &=& { + 4} \\ \\ 6|x - 1| &=& 11 \end{eqnarray} $ Divide both sides by ${6}$ $ \dfrac{6|x - 1|} {{6}} = \dfrac{11} {{6}} $ Simplify: $ |x - 1| = \dfrac{11}{6}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 1 = -\dfrac{11}{6} $ or $ x - 1 = \dfrac{11}{6} $ Solve for the solution where $x - 1$ is negative: $ x - 1 = -\dfrac{11}{6} $ Add ${1}$ to both sides: $ \begin{eqnarray} x - 1 &=& -\dfrac{11}{6} \\ \\ {+ 1} && {+ 1} \\ \\ x &=& -\dfrac{11}{6} + 1 \end{eqnarray} $ Change the ${ + 1}$ to an equivalent fraction with a denominator of $6$ $ x = - \dfrac{11}{6} {+ \dfrac{6}{6}} $ $ x = -\dfrac{5}{6} $ Then calculate the solution where $x - 1$ is positive: $ x - 1 = \dfrac{11}{6} $ Add ${1}$ to both sides: $ \begin{eqnarray} x - 1 &=& \dfrac{11}{6} \\ \\ {+ 1} && {+ 1} \\ \\ x &=& \dfrac{11}{6} + 1 \end{eqnarray} $ Change the ${ + 1}$ to an equivalent fraction with a denominator of $6$ $ x = \dfrac{11}{6} {+ \dfrac{6}{6}} $ $ x = \dfrac{17}{6} $ Thus, the correct answer is $x = -\dfrac{5}{6} $ or $x = \dfrac{17}{6} $.